Problem: Find $\lim_{x\to 2}(9-4x)^{^{\frac{1}{\tan(x-2)}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{e^2}$ (Choice B) B $e^3$ (Choice C) C $\dfrac{1}{e^4}$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=2$ into $(9-4x)^{^{\frac{1}{\tan(x-2)}}}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(9-4x)^{^{\frac{1}{\tan(x-2)}}}$, we will find $\lim_{x\to 2}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 2}y$. $\ln(y) =\dfrac{\ln(9-4x)}{\tan(x-2)}$ Substituting $x=2$ into $\dfrac{\ln(9-4x)}{\tan(x-2)}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 2}\ln(y) \\\\ &=\lim_{x\to 2}\dfrac{\ln(9-4x)}{\tan(x-2)} \\\\ &=\lim_{x\to 2}\dfrac{\dfrac{d}{dx}[\ln(9-4x)]}{\dfrac{d}{dx}[\tan(x-2)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 2}\dfrac{\left(\dfrac{-4}{9-4x}\right)}{\sec^2(x-2)} \\\\ &=\lim_{x\to 2}\dfrac{-4\cos^2(x-2)}{9-4x} \\\\ &=\dfrac{-4}{1} \gray{\text{Substitution}} \\\\ &=-4 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 2}\dfrac{\dfrac{d}{dx}[\ln(9-4x)]}{\dfrac{d}{dx}[\tan(x-2)]}$ actually exists. We found that $\lim_{x\to 2}\ln(y)=-4$, which means $\lim_{x\to 2}y=\dfrac{1}{e^4}$. [Why?] In conclusion, $\lim_{x\to 2}(9-4x)^{^{\frac{1}{\tan(x-2)}}}=\dfrac{1}{e^4}$.